# Current Affairs JEE Main & Advanced

#### Properties of G.P.

(1) If all the terms of a G.P. be multiplied or divided by the same non-zero constant, then it remains a G.P., with the same common ratio.   (2) The reciprocal of the terms of a given G.P. form a G.P. with common ratio as reciprocal of the common ratio of the original G.P.   (3) If each term of a G.P. with common ratio r be raised to the same power k, the resulting sequence also forms a G.P. with common ratio ${{r}^{k}}$.   (4) In a finite G.P., the product of terms equidistant from the beginning and the end is always the same and is equal to the product of the first and last term. i.e., if ${{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,......\,{{a}_{n}}$ be in G.P.   Then ${{a}_{1}}\,{{a}_{n}}={{a}_{2}}\,{{a}_{n-1}}={{a}_{3}}\,{{a}_{n-2}}={{a}_{4}}\,{{a}_{n-3}}=..........={{a}_{r}}\,.\,{{a}_{n-r+1}}$   (5) If the terms of a given G.P. are chosen at regular intervals, then the new sequence so formed also forms a G.P.   (6) If ${{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,.....,\,{{a}_{n}}......$ is a G.P. of non-zero, non-negative terms, then $\log {{a}_{1}},\,\log {{a}_{2}},\,\log {{a}_{3}},\,.....\log {{a}_{n}},\,......$ is an A.P. and vice-versa.   (7) Three non-zero numbers a, b, c are in G.P., iff ${{b}^{2}}=ac$.   (8) If first term of a G.P. of $n$ terms is $a$ and last term is $l,$ then the product of all terms of the G.P. is ${{(al)}^{n/2}}$.   (9) If there be $n$ quantities in G.P. whose common ratio is $r$ and ${{S}_{m}}$ denotes the sum of the first m terms, then the sum of their product taken two by two is $\frac{r}{r+1}\,{{S}_{n}}\,{{S}_{n-1}}$.   (10) If ${{a}^{{{x}_{1}}}},{{a}^{{{x}_{2}}}},{{a}^{{{x}_{3}}}},....,{{a}^{{{x}_{n}}}}$ are in G.P., then ${{x}_{1}},{{x}_{2}},{{x}_{3}},....,{{x}_{n}}$ will be are  in  A.P. ,

#### Definition

A progression is called a harmonic progression (H.P.) if the reciprocals of its terms are in A.P.   Standard form : $\frac{1}{a}+\frac{1}{a+d}+\frac{1}{a+2d}+....$..   Example: The sequence $1,\,\frac{1}{3},\,\frac{1}{5},\,\frac{1}{7},\,\frac{1}{9},...$ is a H.P., because the sequence 1, 3, 5, 7, 9, ….. is an A.P.

#### General Term of an H.P.

If the H.P. be as $\frac{1}{a},\,\frac{1}{a+d},\,\frac{1}{a+2d},\,....$ then corresponding A.P. is $a,\,a+d,\,a+2d,\,.....$   ${{T}_{n}}$ of A.P. is $a+(n-1)\,d$   $\therefore$ ${{T}_{n}}$ of H.P. is $\frac{1}{a+(n-1)\,d}$   In order to solve the question on H.P., we should form the corresponding A.P. Thus, General term :   ${{T}_{n}}=\frac{1}{a+(n-1)\,d}$ or ${{T}_{n}}\text{ of H}\text{.P}\text{.}=\frac{1}{{{T}_{n}}\text{ of A}\text{.P}\text{.}}$.

#### Harmonic Mean

If three or more numbers are in H.P., then the numbers lying between the first and last are called harmonic means (H.M.’s) between them. For example 1, 1/3, 1/5, 1/7, 1/9 are in H.P. So 1/3, 1/5 and 1/7 are three H.M.’s between 1 and 1/9.   Also, if a, H, b are in H.P., then H is called harmonic mean between $a$ and $b$.   (1) Insertion of harmonic means   (i) Single H.M. between $a$ and $b$$=\frac{2ab}{a+b}$.   (ii) H, H.M. of $n$ non-zero numbers ${{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,....,\,{{a}_{n}}$  is given by $\frac{1}{H}=\frac{\frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+.....+\frac{1}{{{a}_{n}}}}{n}$.   (iii) Let $a,\,\,b$ be two given numbers. If $n$ numbers ${{H}_{1}},\,{{H}_{2}},\,......\,{{H}_{n}}$ are inserted between $a$ and $b$ such that the sequence  $a,\,{{H}_{1}},\,{{H}_{2}},\,{{H}_{3}},......\,{{H}_{n}},\,b$ is a H.P., then ${{H}_{1}},\,{{H}_{2}},\,......\,{{H}_{n}}$ are called $n$ harmonic means between $a$ and $b$.   Now, $a,\,{{H}_{1}},\,{{H}_{2}},\,{{H}_{3}},......\,{{H}_{n}},\,b$ are in H.P.   $\Rightarrow$ $\frac{1}{a},\,\frac{1}{{{H}_{1}}},\,\frac{1}{{{H}_{2}}},\,......\frac{1}{{{H}_{n}}},\,\frac{1}{b}$ are in A.P.   Let $D$ be the common difference of this A.P. Then,   $\frac{1}{b}={{(n+2)}^{th}}\text{ term }={{T}_{n+2}}$   $\frac{1}{b}=\frac{1}{a}+(n+1)\,D$$\Rightarrow$$D=\frac{a-b}{(n+1)\,ab}$.   Thus, if $n$ harmonic means are inserted between two given numbers $a$ and $b,$ then the common difference of the corresponding A.P. is given by $D=\frac{a-b}{(n+1)\,ab}$.   Also, $\frac{1}{{{H}_{1}}}=\frac{1}{a}+D$, $\frac{1}{{{H}_{2}}}=\frac{1}{a}+2D$,…….,$\frac{1}{{{H}_{n}}}=\frac{1}{a}+nD$,    where $D=\frac{a-b}{(n+1)\,ab}$.

#### Properties of H.P.

(1) No term of H.P. can be zero.            (2) If H is the H.M. between a and b, then   (i) $\frac{1}{H-a}+\frac{1}{H-b}=\frac{1}{a}+\frac{1}{b}$   (ii) $(H-2a)(H-2b)={{H}^{2}}$        (iii)          $\frac{H+a}{H-a}+\frac{H+b}{H-b}=2$

#### Definition

The combination of arithmetic and geometric progression is called arithmetico-geometric progression.

#### ${{n}^{th}}$term of A.G.P.

If ${{a}_{1}},\,{{a}_{2}},\,{{a}_{3}},\,......,\,{{a}_{n}},\,......$ is an A.P. and ${{b}_{1}},\,{{b}_{2}},\,\,......,\,{{b}_{n}},\,......$ is a G.P., then the sequence ${{a}_{1}}{{b}_{1}},\,{{a}_{2}}{{b}_{2}},\,{{a}_{3}}{{b}_{3}},$$\,......,\,{{a}_{n}}{{b}_{n}},\,.....$ is said to be an arithmetico-geometric sequence.   Thus, the general form of an arithmetico geometric sequence is $a,\,(a+d)\,r,\,(a+2d)\,{{r}^{2}},\,(a+3d)\,{{r}^{3}},\,.....$   From the symmetry we obtain that the nth term of this sequence is $[a+(n-1)\,d]\,{{r}^{n-1}}$.   Also, let $a,\,(a+d)\,r,\,(a+2d)\,{{r}^{2}},\,(a+3d)\,{{r}^{3}},\,.....$be an arithmetico-geometric sequence.   Then, $a+\,(a+d)\,r$$+\,(a+2d)\,{{r}^{2}}+(a+3d)\,{{r}^{3}}+...$ is an arithmetico-geometric series.

#### Sum of A.G.P.

(1) Sum of $n$ terms : The sum of n terms of an arithmetico-geometric sequence $a,\,(a+d)\,r,\,(a+2d)\,{{r}^{2}},\,$$(a+3d)\,{{r}^{3}},\,.....$ is given by  {{S}_{n}}=\left\{ \begin{align}& \frac{a}{1-r}+dr\frac{(1-{{r}^{n-1}})}{{{(1-r)}^{2}}}-\frac{\{a+(n-1)\,d\}{{r}^{n}}}{1-r},\text{ when }r\ne 1 \\& \frac{\text{n}}{\text{2}}[2a+(n-1)\,d],\text{ when }r=1 \\\end{align} \right.\text{ }   (2) Sum of infinite sequence: Let $|r|\,<1$. Then ${{r}^{n}},\,{{r}^{n-1}}\to 0$ as $n\to \infty$ and it can also be shown that $n\,.\,{{r}^{n}}\to 0$ as $n\to \infty$. So, we obtain that ${{S}_{n}}\to \frac{a}{1-r}+\frac{dr}{{{(1-r)}^{2}}}$, as $n\to \infty$¥.   In other words, when  $|r|\,<1$ the sum to infinity of an arithmetico-geometric series is ${{S}_{\infty }}=\frac{a}{1-r}+\frac{dr}{{{(1-r)}^{2}}}$.

#### Method for Finding Sum

This method is applicable for both sum of $n$ terms and sum of infinite number of terms.   First suppose that sum of the series is $S,$ then multiply it by common ratio of the G.P. and subtract. In this way, we shall get a G.P., whose sum can be easily obtained.

#### Method of Difference

If the differences of the successive terms of a series are in A.P. or G.P., we can find ${{n}^{th}}$ term of the series by the following steps :   Step I: Denote the ${{n}^{th}}$ term by ${{T}_{n}}$ and the sum of the series upto $n$ terms by ${{S}_{n}}$.   Step II: Rewrite the given series with each term shifted by one place to the right.   Step III: By subtracting the later series from the former, find ${{T}_{n}}$.   Step IV: From ${{T}_{n}}$, ${{S}_{n}}$ can be found by appropriate summation.   Example : Consider the series 1+ 3 + 6 + 10 + 15 +…..to $n$ terms. Here differences between the successive terms are $63,\text{ }106,\text{ }1510,\text{ }\ldots \ldots .$ i.e.,  2, 3, 4, 5,…… which are in A.P. This difference could be in G.P. also. Now let us find its sum   $S=1+3+6+10+15+.....+{{T}_{n-1}}+{{T}_{n}}$   $S=\,\,\,\,\,\,\,\,\,1+3+6+10+..........+{{T}_{n-1}}+{{T}_{n}}$   Subtracting, we get   $0=1+2+3+4+5+.........+({{T}_{n}}-{{T}_{n-1}})-{{T}_{n}}$   $\Rightarrow$      ${{T}_{n}}=1+2+3+4+.........$to $n$ terms.   $\Rightarrow$      ${{T}_{n}}=\frac{1}{2}n(n+1)$     $\therefore$ ${{S}_{n}}=\Sigma {{T}_{n}}=\frac{1}{2}[\Sigma {{n}^{2}}+\Sigma n]$   = $\frac{1}{2}\left[ \frac{n(n+1)\,(2n+1)}{6}+\frac{n\,(n+1)}{2} \right]$ = $\frac{n\,(n+1)\,(n+2)}{6}$.

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